Solution a is 2 mmol/l urea, and solution b is 1mmol/l nacl. assume that g nacl = 1.85. are the 2 solutions isosmotic?

1 answer:

8 0

Soution A has a higher osmolarity than solution B and is hyperosmotic.
Solution be is hyposmotic. Neither are isosmotic.

You might be interested in

How many centigrams are in 0.13 oz

Paha777 [63]

Answer:

I think the answer is 368.544 centigrams

Explanation:

I hope I helped

5 0

3 years ago

Read 2 more answers

Rank the following elements by effective nuclear charge, Zeff, for a valence electron. F LI Be B N

Stels [109]

Answer:

Rank in increasing order of effective nuclear charge:

Li < Be < B < N < F

Explanation:

This explains the meaning of effective nuclear charge, Zeff, how to determine it, and the calculations for a valence electron of each of the five given elements: F, Li, Be, B, and N.

1) Effective nuclear charge definitions

While the total positive charge of the atom nucleus (Z) is equal to the number of protons, the electrons farther away from the nucleus experience an effective nuclear charge (Zeff) less than the total nuclear charge, due to the fact that electrons in between the nucleus and the outer electrons partially cancel the atraction from the nucleus.

Such effect on on a valence electron is estimated as the atomic number less the number of electrons closer to the nucleus than the electron whose effective nuclear charge is being determined: Zeff = Z - S.

2) Z eff for a F valence electron:

F's atomic number: Z = 9
Total number of electrons: 9 (same numer of protons)
Period: 17 (search in the periodic table or do the electron configuration)
Number of valence electrons: 7 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 9 - 7 = 2
Zeff = Z - S = 9 - 2 = 7

3) Z eff for a Li valence eletron:

Li's atomic number: Z = 3
Total number of electrons: 3 (same number of protons)
Period: 1 (search on the periodic table or do the electron configuration)
Number of valence electrons: 1 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 3 - 1 = 2
Z eff = Z - S = 3 - 2 = 1.

4) Z eff for a Be valence eletron:

Be's atomic number: Z = 4
Total number of electrons: 4 (same number of protons)
Period: 2 (search on the periodic table or do the electron configuration)
Number of valence electrons: 2 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 4 - 2 = 2
Z eff = Z - S = 4 - 2 = 2

5) Z eff for a B valence eletron:

B's atomic number: Z = 5
Total number of electrons: 5 (same number of protons)
Period: 13 (search on the periodic table or do the electron configuration)
Number of valence electrons: 3 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 5 - 3 = 2
Z eff = Z - S = 5 - 2 = 3

6) Z eff for a N valence eletron:

N's atomic number: Z = 7
Total number of electrons: 7 (same number of protons)
Period: 15 (search on the periodic table or do the electron configuration)
Number of valence electrons: 5 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 7 - 5 = 2
Z eff = Z - S = 7 - 2 = 5

7) Summary (order):

Atom Zeff for a valence electron